Question: Simplify and expand the following expression: $ \dfrac{1}{3q - 24}- \dfrac{5}{2q - 8}+ \dfrac{q}{q^2 - 12q + 32} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the first term: $ \dfrac{1}{3q - 24} = \dfrac{1}{3(q - 8)}$ We can factor a $2$ out of denominator in the second term: $ \dfrac{5}{2q - 8} = \dfrac{5}{2(q - 4)}$ We can factor the quadratic in the third term: $ \dfrac{q}{q^2 - 12q + 32} = \dfrac{q}{(q - 8)(q - 4)}$ Now we have: $ \dfrac{1}{3(q - 8)}- \dfrac{5}{2(q - 4)}+ \dfrac{q}{(q - 8)(q - 4)} $ The least common multiple of the denominators is: $ 6(q - 8)(q - 4)$ In order to get the first term over $6(q - 8)(q - 4)$ , multiply by $\dfrac{2(q - 4)}{2(q - 4)}$ $ \dfrac{1}{3(q - 8)} \times \dfrac{2(q - 4)}{2(q - 4)} = \dfrac{2(q - 4)}{6(q - 8)(q - 4)} $ In order to get the second term over $6(q - 8)(q - 4)$ , multiply by $\dfrac{3(q - 8)}{3(q - 8)}$ $ \dfrac{5}{2(q - 4)} \times \dfrac{3(q - 8)}{3(q - 8)} = \dfrac{15(q - 8)}{6(q - 8)(q - 4)} $ In order to get the third term over $6(q - 8)(q - 4)$ , multiply by $\dfrac{6}{6}$ $ \dfrac{q}{(q - 8)(q - 4)} \times \dfrac{6}{6} = \dfrac{6q}{6(q - 8)(q - 4)} $ Now we have: $ \dfrac{2(q - 4)}{6(q - 8)(q - 4)} - \dfrac{15(q - 8)}{6(q - 8)(q - 4)} + \dfrac{6q}{6(q - 8)(q - 4)} $ $ = \dfrac{ 2(q - 4) - 15(q - 8) + 6q} {6(q - 8)(q - 4)} $ Expand: $ = \dfrac{2q - 8 - 15q + 120 + 6q}{6q^2 - 72q + 192} $ $ = \dfrac{-7q + 112}{6q^2 - 72q + 192}$